Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
const int K = 17;
const int M = 10;
int n, m, q;
vector<int> adj[N];
int depth[N];
int P[N][K];
vector<int> self[N];
vector<int> data[N][K];
void dfs(int u, int p, int d) {
  depth[u] = d;
  for (int v : adj[u]) {
    if (v == p) continue;
    dfs(v, u, d + 1);
    P[v][0] = u;
  }
}
void Merge(vector<int>& a, vector<int>& b) {
  vector<int> c;
  int n = a.size(), m = b.size();
  int i = 0, j = 0;
  while (c.size() < M && (i < n || j < m)) {
    if (i == n) {
      c.push_back(b[j++]);
    } else if (j == m) {
      c.push_back(a[i++]);
    } else if (a[i] < b[j]) {
      c.push_back(a[i++]);
    } else if (a[i] > b[j]) {
      c.push_back(b[j++]);
    } else {
      c.push_back(a[i++]);
      ++j;
    }
  }
  a = c;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m >> q;
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  for (int i = 1; i <= m; ++i) {
    int a;
    cin >> a;
    self[a].push_back(i);
  }
  for (int i = 1; i <= n; ++i) {
    P[i][0] = -1;
  }
  dfs(1, -1, 0);
  for (int i = 1; i <= n; ++i) {
    data[i][0] = self[i];
    if (P[i][0] != -1) {
      Merge(data[i][0], self[P[i][0]]);
    }
  }
  for (int j = 1; j < K; ++j) {
    for (int i = 1; i <= n; ++i) {
      int k = P[i][j - 1];
      if (k == -1) {
        P[i][j] = -1;
      } else {
        P[i][j] = P[k][j - 1];
        data[i][j] = data[i][j - 1];
        Merge(data[i][j], data[k][j - 1]);
      }
    }
  }
  while (q--) {
    int a, b, sz;
    cin >> a >> b >> sz;
    vector<int> res;
    if (depth[a] < depth[b]) swap(a, b);
    Merge(res, self[a]);
    Merge(res, self[b]);
    for (int j = K - 1; j >= 0; --j) {
      if (depth[a] - (1 << j) >= depth[b]) {
        Merge(res, data[a][j]);
        a = P[a][j];
      }
    }
    if (a != b) {
      for (int j = K - 1; j >= 0; --j) {
        if (P[a][j] != -1 && P[a][j] != P[b][j]) {
          Merge(res, data[a][j]);
          Merge(res, data[b][j]);
          a = P[a][j];
          b = P[b][j];
        }
      }
      Merge(res, self[P[a][0]]);
    }
    sz = min(sz, (int)res.size());
    cout << sz << ' ';
    for (int i = 0; i < sz; ++i) {
      cout << res[i] << ' ';
    }
    cout << '\n';
  }
}